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A342326
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a(n) is the smallest nonnegative integer that can be written as a sum of two distinct nonzero triangular numbers in exactly n ways or -1 if no such integer exists.
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2
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0, 4, 16, 81, 471, 2031, 1381, 11781, 6906, 17956, 34531, 123256, 40056, 305256, 863281, 448906, 200281, 1957231, 520731, 10563906, 1001406, 11222656, 7631406, 3454506, 1482081, 75865156, 7172606106, 8852431, 25035156, 334020781, 13018281, 38531031, 7410406, 7014160156
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OFFSET
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0,2
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COMMENTS
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Conjecture: This sequence has a positive a(n) for every positive integer n, and each sequence in the infinite indexed family, of which this sequence offers the initial terms, is infinite, as well.
a(40) = 37052031, a(45) = 221310781, a(48) = 60765331, a(39) <= 2782318906, a(42) <= 325457031, a(47) <= 927577056, a(50) <= 2200089531, a(54) <= 327539956, a(56) <= 926300781, a(60) <= 481676406, a(63) <= 4598740656, a(64) <= 303826656, a(71) <= 4579579956, a(72) <= 789949306, a(80) <= 1519133281, a(96) <= 3220562556. Terms for n <= 96 not listed here and terms for which only upper bounds are known are >= 3*10^8.
Is a(n) == 6 (mod 25) for n >= 5? It holds for all terms known to date.
The triangular numbers mod 25 are periodic with period 25. Constructing all 25*25 = 625 sums of two distinct triangular numbers mod 25 gives 65 cases for 6 (mod 25). The second largest occurs 40 times. (End)
a(47) = 550240551, a(59) = 7629645156, a(67) = 6418012656, a(81) = 9498658731, a(90) = 8188498906. All upper bounds listed in the above comments for n other than 47 are the exact values of a(n). For all n for which no value is listed here or above, a(n) > 10^10 (or a(n) = -1). - Jon E. Schoenfield, Mar 09 2021
a(44) = 15646972656. For n<=51, all terms not mentioned here or above, a(n) >= 6.5*10^10 (or a(n) = -1).
a(47) == 1 (mod 25) and a(95) = 47652012541 == 16 (mod 25). Thus the answer to Corneth's question is 'No'. (End)
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LINKS
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FORMULA
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EXAMPLE
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a(1) = 4 = 1 + 3;
a(2) = 16 = 1 + 15 = 6 + 10;
a(3) = 81 = 3 + 78 = 15 + 66 = 36 + 45.
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MATHEMATICA
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r = 125000; (* generates the first 12 terms of the sequence *)
lst = Table[0, {r}];
lim = Floor[Sqrt[2r]];
Do[ num = (i^2 + i)/2 + (j^2 + j)/2;
If[num <= r, lst[[num]]++], {i, lim}, {j, i - 1}];
First /@ (Flatten@Position[lst, #] & /@ Range[Max[lst]])
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PROG
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(PARI) upto(n) = {my(v = vector(n)); res = vector(10); for(i = 1, (sqrtint(8*n + 1)-1)\2, bi = binomial(i + 1, 2); for(j = i+1, (sqrtint(8*(n - bi))-1)\2, v[bi + binomial(j+1, 2)]++ ) ); for(i = 1, #v, if(v[i] > 0, if(v[i] > #res, res = concat(res, vector(v[i] - #res)); ); if(res[v[i]] == 0, res[v[i]] = i ) ) ); concat(0, res) } \\ David A. Corneth, Mar 08 2021
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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