

A020756


Numbers that are the sum of two triangular numbers.


20



0, 1, 2, 3, 4, 6, 7, 9, 10, 11, 12, 13, 15, 16, 18, 20, 21, 22, 24, 25, 27, 28, 29, 30, 31, 34, 36, 37, 38, 39, 42, 43, 45, 46, 48, 49, 51, 55, 56, 57, 58, 60, 61, 64, 65, 66, 67, 69, 70, 72, 73, 76, 78, 79, 81, 83, 84, 87, 88, 90, 91, 92, 93, 94, 97, 99, 100, 101, 102, 105, 106, 108
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OFFSET

1,3


COMMENTS

The possible sums of a square and a promic, i.e., x^2+n(n+1), e.g., 3^2 + 2*3 = 9 + 6 = 15 is present.  Jon Perry, May 28 2003


LINKS



FORMULA

Numbers n such that 4n+1 is the sum of two squares, i.e. such that 4n+1 is in A001481. Hence n is a member if and only if 4n+1 = odd square * product of distinct primes of form 4k+1. (Fred Helenius and others, Dec 18 2004)
Equivalently, we may say that a positive integer n can be partitioned into a sum of two triangular numbers if and only if every 4 k + 3 prime factor in the canonical form of 4 n + 1 occurs with an even exponent.  Ant King, Nov 29 2010
Also, the values of n for which 8n+2 can be partitioned into a sum of two squares of natural numbers.  Ant King, Nov 29 2010
Closed under the operation f(x, y) = 4*x*y + x + y.


MATHEMATICA

q[k_] := If[! Head[Reduce[m (m + 1) + n (n + 1) == 2 k && 0 <= m && 0 <= n, {m, n}, Integers]] === Symbol, k, {}]; DeleteCases[Table[q[i], {i, 0, 108}], {}] (* Ant King, Nov 29 2010 *)
Take[Union[Total/@Tuples[Accumulate[Range[0, 20]], 2]], 80] (* Harvey P. Dale, May 02 2012 *)


PROG

(PARI) v=vector(200); vc=0; for (x=0, 10, for (y=0, 10, v[vc++ ]=x^2+y*(y+1))); v=vecsort(v); v
(PARI) is(n)=my(f=factor(4*n+1)); for(i=1, #f~, if(f[i, 1]%4==3 && f[i, 2]%2, return(0))); 1 \\ Charles R Greathouse IV, Jul 05 2013
(Haskell)
a020756 n = a020756_list !! (n1)
a020756_list = filter ((> 0) . a052343) [0..]


CROSSREFS



KEYWORD

nonn,nice


AUTHOR



EXTENSIONS



STATUS

approved



