%I #92 Jun 09 2024 09:02:38
%S 2,4,7,13,22,40,67,121,202,364,607,1093,1822,3280,5467,9841,16402,
%T 29524,49207,88573,147622,265720,442867,797161,1328602,2391484,
%U 3985807,7174453,11957422,21523360,35872267,64570081,107616802,193710244,322850407,581130733,968551222,1743392200
%N Least number m for which there exists some positive k < m where the sum of the integers from k + 1 to m inclusive is an n-th power > 1.
%C With triangular sum T(i) = i*(i+1)/2 = Sum_{j=1..i} i, the aim is T(m) - T(k) = b^n for some b. T(m) - T(k) = (m+k+1)*(m-k)/2 so if m-k is odd then use m-k = b^x to eliminate k in the other factor (m+k+1)/2 = b^(n-x), so that m = b^(n-x) + (b^x - 1)/2. If instead m+k+1 is odd then the resulting expression is the same. This form is an integer and minimized at b=3 and x = ceiling(n/2). - _Kevin Ryde_, May 18 2024
%H Paolo Xausa, <a href="/A372782/b372782.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (1,3,-3).
%F a(n) = A087503(n-1) + 1.
%F a(n) = 3*a(n-2) + 1.
%F G.f.: x*(2 + 2*x - 3*x^2)/((1 - x)*(1 - 3*x^2)). - _Stefano Spezia_, May 18 2024
%e a(2) = 4 because the sum of all integers from 3 + 1 to 4 inclusive is 4 = 2^2, a square.
%e a(3) = 7 as we have 2 + 3 + 4 + 5 + 6 + 7 = 27 = 3^3, i.e., m = 7 and k = 1. - _David A. Corneth_, May 15 2024
%t LinearRecurrence[{1, 3, -3}, {2, 4, 7}, 50] (* _Paolo Xausa_, Jun 09 2024 *)
%o (PARI) isok(m, n) = my(s = m*(m+1)/2); for (k=1, m-1, s -= k; if (ispower(s, n), return(k)););
%o a(n) = my(m=1); while (! isok(m, n), m++); m; \\ _Michel Marcus_, May 16 2024
%o (PARI) a(n) = {
%o my(res, c, l, u);
%o res = 2^n; c = 8*3^n;
%o l = (sqrt(c) - 1)\2; u = res;
%o for(i = l, u,
%o if(issquare(4*(i + 1)*i + 1 - c),
%o return(i);
%o )
%o );
%o return(2^n)
%o } \\ _David A. Corneth_, May 16 2024
%Y Cf. A002452, A052993, A087503, A372631.
%K nonn,easy
%O 1,1
%A _Jean-Marc Rebert_, May 14 2024
%E More terms from _David A. Corneth_, May 16 2024