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 A226789 Triangular numbers obtained as the concatenation of n+1 and n. 3
 10, 21, 26519722651971, 33388573338856, 69954026995401, 80863378086336 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS There are only six terms less than 10^20. LINKS Table of n, a(n) for n=1..6. EXAMPLE 26519722651971 is the concatenation of 2651972 and 2651971 and a triangular number, because 26519722651971 = 7282818*7282819/2. MATHEMATICA TriangularQ[n_] := IntegerQ[Sqrt[1 + 8*n]]; t = {}; Do[s = FromDigits[Join[IntegerDigits[n+1], IntegerDigits[n]]]; If[TriangularQ[s], AppendTo[t, s]], {n, 100000}]; t (* T. D. Noe, Jun 18 2013 *) PROG (PARI) concatint(a, b)=eval(concat(Str(a), Str(b))) istriang(x)=issquare(8*x+1) {for(n=1, 10^7, a=concatint(n+1, n); if(istriang(a), print(a)))} (Python) from math import isqrt def istri(n): t = 8*n+1; return isqrt(t)**2 == t def afind(klimit, kstart=0): strk = "0" for k in range(kstart, klimit+1): strkp1 = str(k+1) t = int(strkp1 + strk) if istri(t): print(t, end=", ") strk = strkp1 afind(81*10**5) # Michael S. Branicky, Oct 21 2021 (Python) # alternate version def isconcat(n): if n < 10: return False s = str(n) mid = (len(s)+1)//2 lft, rgt = int(s[:mid]), int(s[mid:]) return lft - 1 == rgt def afind(tlimit, tstart=0): for t in range(tstart, tlimit+1): trit = t*(t+1)//2 if isconcat(trit): print(trit, end=", ") afind(13*10**6) # Michael S. Branicky, Oct 21 2021 CROSSREFS Cf. A003098, A068899, A226742, A226772, A226788. Sequence in context: A109326 A369120 A080454 * A110367 A104865 A063555 Adjacent sequences: A226786 A226787 A226788 * A226790 A226791 A226792 KEYWORD nonn,base,more AUTHOR Antonio Roldán, Jun 18 2013 STATUS approved

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Last modified June 16 08:45 EDT 2024. Contains 373424 sequences. (Running on oeis4.)