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A261127
Triangular numbers t such that (sum of digits of t) + (product of digits of t) is a triangular number.
1
0, 3, 10, 105, 120, 136, 190, 210, 300, 406, 703, 780, 820, 1081, 1128, 1431, 1540, 1653, 1770, 1891, 1953, 2080, 2211, 2628, 2701, 2850, 3003, 3160, 3403, 3570, 4560, 4656, 5050, 5460, 7021, 7260, 7503, 8646, 8911, 9453, 10011, 10153, 11026, 12403, 14028, 15400
OFFSET
1,2
COMMENTS
All the terms in this sequence are triangular, and hence 0 or 1 (mod 3).
LINKS
EXAMPLE
a(6) = 136 = 16 * (16+1) / 2, that is triangular number. (1+3+6) + (1*3*6) = 28, which is 7th triangular number.
a(15) = 1128 = 47 * (47+1) / 2, that is triangular number. (1+1+2+8) + (1*1*2*8) = 28, which is 7th triangular number.
MAPLE
with(numtheory): A261127:= proc() local a, k, t; t:=n*(n+1)/2; a:= (add(d, d=convert(t, base, 10)) + mul(d, d=convert(t, base, 10))); k:=(-1 + sqrt(8*a + 1))/2; if k=floor(k) then RETURN (t); fi; end: seq(A261127 (), n=0..300);
MATHEMATICA
A261127 = {}; Do[t = n*(n + 1)/2; k = Plus @@ IntegerDigits[t] + Times @@ IntegerDigits[t]; If[IntegerQ[( -1 + Sqrt[8*k + 1])/2], AppendTo[A261127, t]], {n, 0, 1000}]; A261127
PROG
(PARI) for(n =0, 500, t = n*(n+1)/2; k = (sumdigits(t)); d = digits (t); p = prod(i = 1, #d, d[i]); s = k+p; if(ispolygonal(s, 3), print1(t, ", ")));
(Magma) [n*(n+1) div 2: n in [0..100] | IsSquare(8*k+1) where k is (&+Intseq(n*(n+1) div 2) + &*Intseq(n*(n+1) div 2))];
CROSSREFS
KEYWORD
nonn,base
AUTHOR
K. D. Bajpai, Aug 09 2015
STATUS
approved