A306122 Numbers that are product of a second hexagonal number (A014105) and a square pyramidal numbers (A000330) in at least two ways.

0, 105, 300, 855, 1155, 2940, 13860, 14700, 17850, 20790, 22230, 27300, 33930, 70125, 73920, 87780, 114400, 116025, 135135, 145530, 157080, 195000, 213150, 235290, 304590, 347655, 381150, 431340, 451044, 471975, 566580, 632700, 764400, 796950, 942480, 950040

Offset: 1

Author

Geoffrey B. Campbell and M. F. Hasler, Jul 03 2018

Keywords

• nonn

Comments

We have A000330(n) = 1 + 2^2 + ... + n^2 and A014105(m) = 0^2 - 1^2 + 2^2 -+ ... + (2m)^2, so the terms of this sequence are the numbers that are a product, in at least two ways, of a partial sum of squares times a (positive) partial sum of squares with alternating signs (with + for even terms; cf. A306121 for the opposite convention).

The initial a(1) = 0 is added for completeness.

Below 10^8, the number 17850 is the only one to have four representations of the given form, and 6347250 is the only one to have exactly three.

Links

• Geoffrey Campbell, Integer solutions of (1²-2²+3²-...+(2a-1)²) × (1²+2²+3²+...+b²) = (1²-2²+3²-...+(2c-1)²) × (1²+2²+3²+...+d²) where a ≠ c and b ≠ d, Number Theory group on LinkedIn, June 2018.

Crossrefs

Cf. A000330, A014105, A306121.

Programs

PARI

{my(L=10^6,A14105(a)=a*(2*a+1),A330(b)=(b+1)*b*(2*b+1)/6,A=S=[]); for(b=1, sqrtnint(L\A14105(1)\3,3), for(a=1,oo, if( setsearch(S,t=A14105(a)*A330(b)), A=setunion(A,[t]), t>L&&next(2); S=setunion(S,[t]))));A}