login
A227115
Powers but not squares which are sum of consecutive composites less than 10^7 ordered according to the proximity of the first composite of the sum to the first composite: 4.
0
27, 10077696, 128, 32768, 8, 27, 1000, 1728, 5088448, 690807104, 27, 32, 512, 2048, 512, 6859, 4913, 243, 405224, 125, 3125, 2744, 98611128, 27000, 314432, 216, 1728, 1889568, 243, 2744, 512, 4913000
OFFSET
1,1
COMMENTS
There are other informative data for each term of the sequence. They are (b,l,k) where b is the base to an odd power, l is the number of consecutive composites added, and k indicates the k-th composite c(k) from where the sums begin: (3,4,1), (6,4151,1), (2,10,2), (2,222,2), (2,1,3), (3,3,3), (10,30,7), (12,42,7), (172,2931,7), (884,35029,9), (3,1,17), (2,1,20), (2,13,20), (2,36,22), (2,12,23), (19,79,24), (17,59,31), (3,4,41), (74,772,42), (5,2,43), (5,37,43), (14,33,44), (462,13093,46), (30,162,47), (68,668,48), (6,3,50), (12,20,53), (18,1723,56), (3,3,57), (14,28,58), (2,6,59), (170,2827,60).
EXAMPLE
We denote the n-th composite as c(n). Some of the odd powers are the sum of consecutive composites in several ways: 27 = 3^3 = c(1)+c(2)+c(3)+c(4) = c(3)+c(4)+c(5) = c(17) = 4 + 6 + 8 + 9 = 8 + 9 + 10. 243 = 3^5 = c(189) = c(90)+c(91) = c(57)+c(59)+c(59) = c(41)+c(42)+c(43)+c(44) = 121 + 122 = 80 + 81 + 82 = 58 + 60 + 62 + 63. 1000 = 10^3 is sum of 30 consecutive composites beginning with c(7) = 14. 1728 = 12^3 = Ramanujan taxicab minus 1 is sum of 42 consecutive composites beginning with c(7) = 14 and of 20 consecutive composites beginning with c(53) = 75.
PROG
(PARI): n1=10^7; v=vector(n1); i=0; for(a=2, n1, if(isprime(a), next, i++; v[i]=a)); for(b=1, 60, k=0; for(j=b, i, k=k+v[j]; if(ispower(k, , &n)&ispower(k)%2==1, print1([k, n, ispower(k), j-b+1, b], " "))))
CROSSREFS
KEYWORD
nonn,less
AUTHOR
Robin Garcia, Jul 04 2013
STATUS
approved