A342326 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.

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

Offset: 0

Author

Robert G. Root, Mar 08 2021

Keywords

• nonn

Comments

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.

From David A. Corneth, Mar 08 2021: (Start)

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

From Martin Ehrenstein, Mar 09 2021: (Start)

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)

Examples

			a(1) =  4 = 1 +  3;
a(2) = 16 = 1 + 15 =  6 + 10;
a(3) = 81 = 3 + 78 = 15 + 66 = 36 + 45.
		

Links

• Martin Ehrenstein, Table of n, a(n) for n = 0..37

Crossrefs

Cf. A000217, A260647, A051533, A262749, A265140, A265134, A265136, A265138, A307597.

Programs

Mathematica

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]])

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

Formula

a(n) = min { m >= 0 : A307597(m) = n }. - Alois P. Heinz, Mar 08 2021

Extensions

a(13)-a(18) from Alois P. Heinz, Mar 08 2021

a(19)-a(25) from David A. Corneth, Mar 08 2021

a(26)-a(33) from Jon E. Schoenfield, Mar 09 2021 (some terms first found by David A. Corneth)