Which tetrahedral numbers, i.e., numbers of the form (b*(b+1)*(b+2))/6, are also palindromic, meaning their digits read the same forwards and backwards? For example, Patrick De Geest noted that b=336 gives the decimal palindrome 6378736. Of course, if we don't restrict ourselves to decimal representations we can find many such numbers. The base-9 representation of the tetrahedral number k(k+1)(k+2)/6 is palindromic for the following values of k (shown in decimal): k k(k+1)(k+2)/6 (in base 9) ------ ----------------------------- 1 1 2 4 3 11 4 22 12 444 13 555 120 488884 588 71066017 1093 505555505 88573 505055555550505 Needless to say, the "tetrahedral" numbers are just the binomial coefficients C(k,3). The base-3 representations of C(k,3) are palindromic for the following values of k k C(k,3) (in base 3) ------ ----------------------------- 1 1 2 11 3 101 4 202 6 2002 12 111111 14 202202 39 112121211 120 112222222211 392 201000222000102 496 1102111111112011 Similarly we have C(1105,3) = 1534114351 in base 8 C(521,3) = 1122123212211 in base 4 C(1166,3) = 6364334636 in base 7 C(82332,3) = 12 5 9 13 18 18 13 9 5 12 in base 27 There is also a tetrahedral number C(k,3) whose base-B representation is palindromic where both k and B are perfect squares (k=441, B=16). Of course, every positive integer has a palindromic representation in SOME base. If we let f(n) denote the smallest base relative to which n is palindromic, then clearly f(n) is no greater than n-1, because every number n has the palindromic form '11' in the base (n-1). Interestingly, for almost all integers n the value of f(n) is actually quite small, as can be gathered from the table below: n f(n) n f(n) n f(n) n f(n) n f(n) --- ---- --- ---- --- ---- --- ---- --- ---- 1 2 21 2 41 5 61 6 81 8 2 3 22 10 42 4 62 5 82 3 3 2 23 3 43 6 63 2 83 5 4 3 24 5 44 10 64 7 84 11 5 2 25 4 45 2 65 2 85 2 6 5 26 3 46 4 66 10 86 6 7 2 27 2 47 46 67 5 87 28 8 3 28 3 48 7 68 3 88 5 9 2 29 4 49 6 69 22 89 8 10 3 30 9 50 7 70 9 90 14 11 10 31 2 51 2 71 7 91 3 12 5 32 7 52 3 72 5 92 6 13 3 33 2 53 52 73 2 93 2 14 6 34 4 54 8 74 6 94 46 15 2 35 6 55 4 75 14 95 18 16 3 36 5 56 3 76 18 96 11 17 2 37 6 57 5 77 10 97 8 18 5 38 4 58 28 78 5 98 5 19 18 39 12 59 4 79 78 99 2 20 3 40 3 60 9 80 3 100 3 The numbers for which f(n) = n-1 are 3, 4, 6, 11, 19, 47, 53, 103, 137, 139, 149, 163, 167, 179, 223, 263, 269, 283, 293, 311, 317, 347, 359, 367, 389, 439, 491, 563, 569, 593, 607, 659, 739, 827, 853, 877 977, 983,... These might be called "the non-palindromic numbers" because they are the only numbers n that are NOT palindromic in ANY base from 2 to n-2. Proposition: If f(n) = n-1 for n > 6 then n is a prime. Proof: If n=ab with a < (b-1) and b > 2 we have n=a(b-1)+a, so n has the palindromic form 'aa' when expressed in the base (b-1). This covers all composites greater than 6 except for squared primes, but for squares we have a^2 = (a-1)^2 + 2(a-1) + 1, so every square n > 4 has the palindromic form 121 in the base sqrt(n)-1. Done. The ratio of non-palindromic primes to all primes drops as the range increases. For example, of the 1229 primes less than 10000 we find that 218 are non-palindromic, which is about 1 out of every 5.6. Of the 9295 primes less than 100000 we find only 1199 are non-palindromic, which is only about 1 out of every 7.75. At the opposite extreme, there are primes that are already palindromic in the base 2, although these are even more scarce than non-palindromic primes. The first few are 3, 5, 7, 17, 31, 73, 107, 127, 257, 313, 443, 1193, 1453, ... Returning to the function f(n) for general values of n (not restricted to primes), it's clear that f(n), n=1,2,3,... fluctuates between 2 and n-1, but I think the average value of f(n) for all n from 1 to infinity is polynomial in n. Specifically it appears that f(n)_average = n^(1/c) where c is a constant approximately equal to 2.68. In other words, I conjecture that 1 n ln(k) lim --- SUM ------- = c n->inf n k=1 ln(f(k)) but I don't know how to prove that this limit actually exists. Anyway, it's also interesting to observe the values of f(n) when n is a prime power. For example, powers of 2 have the following values of f(n): n f(n) representation of n in the base f(n) --- ---- ------------------------------------ 2 3 2 4 3 1 1 8 3 2 2 16 3 1 2 1 32 7 4 4 64 7 1 2 1 128 7 2 4 2 256 15 1 2 1 512 7 1 3 3 1 1024 7 2 6 6 2 2048 31 2 4 2 4096 7 1 4 6 4 1 8192 15 2 6 6 2 16384 15 4 12 12 4 In general it appears that the min-base palindromic representation of 2^m is a multiple of a binomial expansion, i.e., 2^m = 2^r [(2^s -1) + 1]^t Obviously we have m = st + r. The values of r,s,t for the first several m are tabulated below: m r s t m r s t m r s t - - - - - - - - - - - - 1 1 2 0 11 1 5 2 21 1 5 4 2 0 2 1 12 0 3 4 22 2 5 4 3 1 2 1 13 1 4 3 23 2 7 3 4 0 2 2 14 2 4 3 24 0 6 4 5 2 3 1 15 0 5 3 25 0 5 5 6 0 3 2 16 0 4 4 26 1 5 5 7 1 3 2 17 1 4 4 27 3 6 4 8 0 4 2 18 3 5 3 28 0 7 4 9 0 3 3 19 1 6 3 29 1 7 4 10 1 3 3 20 0 4 5 30 0 5 6 In each case the triple (r,s,t) is such that s is minimized under the constraint that / (2^r) * C(t,t/2) if t is even 2^s - 1 > ( \ (2^r) * C(t-1,(t-1)/2) if t is odd The min-base palindromic representation of other prime powers also tend to be multiples of binomial expansions, but not always. For example, the min-base representation of 3^7 is [3 19 3] in the base 24. Also, the min-base representation of 7^6 is [1 2 3 2 1] in the base 18. This raises the question of whether the min-base representation for powers of 2 is always of the binomial form. Can anyone supply a proof or counter-example?

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