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 A225844 Least k>0 such that triangular(n) + k*(k+1) is a triangular number. 0
 2, 1, 3, 5, 7, 2, 11, 13, 5, 17, 19, 3, 6, 25, 27, 9, 31, 33, 35, 4, 9, 41, 8, 45, 47, 10, 14, 53, 9, 5, 59, 61, 21, 18, 67, 69, 21, 73, 75, 14, 22, 6, 11, 13, 87, 15, 91, 26, 20, 34, 12, 101, 26, 105, 30, 7, 20, 33, 115, 117, 119, 34, 21, 125, 37, 129, 29, 133, 14, 137 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,1 COMMENTS For n>0, a(n) <= 2*n-1, because n*(n+1)/2 + (2*n-1)*2*n =  (9*n^2 - 3*n)/2 = 3*n*(3*n-1)/2 = triangular(3*n-1). The subsequence with terms less than 2*n-1 begins: 2, 5, 3, 6, 9, 4, 9, 8, 10, 14, 9, 5, 21, 18, 21, 14, 22, 6, 11, 13, 15, ... The sequence of n's such that a(n) < 2*n-1 begins: 5, 8, 11, 12, 15, 19, 20, 22, 25, 26, ... LINKS MATHEMATICA lktrno[n_]:=Module[{t=(n(n+1))/2, k=1}, While[!IntegerQ[(Sqrt[ 8(t+k(k+1))+1]-1)/2], k++]; k]; Array[lktrno, 70, 0] (* Harvey P. Dale, Aug 19 2014 *) PROG (Python) def isTriangular(a):     sr = 1L << (long.bit_length(long(a)) >> 1)     a += a     while a < sr*(sr+1):  sr>>=1     b = sr>>1     while b:       s = sr+b       if a >= s*(s+1):  sr = s       b>>=1     return (a==sr*(sr+1)) n = tn = 0 while 1: for m in range(1, 1000000000):     if isTriangular(tn + m*(m+1)): break   print str(m)+', ',   n += 1   tn += n (PARI) a(n)=for(k=1, 2*n, t=n*(n+1)/2+k*(k+1); x=sqrtint(2*t); if(t==x*(x+1)/2, return(k))) /* from Ralf Stephan */ CROSSREFS Cf. A000217, A002378, A082183. Cf. A101157 (least k>0 such that triangular(n) + k^2 is a triangular number). Sequence in context: A275705 A217036 A127201 * A006769 A075643 A076074 Adjacent sequences:  A225841 A225842 A225843 * A225845 A225846 A225847 KEYWORD nonn AUTHOR Alex Ratushnyak, May 17 2013 STATUS approved

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Last modified June 4 10:51 EDT 2020. Contains 334825 sequences. (Running on oeis4.)