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a(n) = Product_{i=1..n} p(i) - p(n+1)^3, where p(i) is the i-th prime.
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%I #16 Nov 06 2024 10:19:39

%S -25,-119,-313,-1121,113,25117,503651,9687523,223068481,6469663439,

%T 200560439477,7420738065889,304250263447703,13082761331566207,

%U 614889782588342533,32589158477189839351,1922760350154212412089,117288381359406970682507,7858321551080267055521179,557940830126698960967026373

%N a(n) = Product_{i=1..n} p(i) - p(n+1)^3, where p(i) is the i-th prime.

%C It is known that a(n) >= 0 for n >= 5.

%D S. W. Golomb, Elementary Problem E3137, Amer. Math. Monthly, Proposed 93 (1986), p. 215; Solution and Editorial Comments, 94 (1987), 883-884.

%H Paolo Xausa, <a href="/A360511/b360511.txt">Table of n, a(n) for n = 1..300</a>

%t FoldList[Times, Most[#]] - Rest[#]^3 & [Prime[Range[25]]] (* _Paolo Xausa_, Nov 06 2024 *)

%o (Python)

%o from sympy import prime, primorial

%o def A360511(n): return primorial(n)-prime(n+1)**3 # _Chai Wah Wu_, Feb 24 2023

%Y Cf. A062234, A064819, A360510.

%K sign

%O 1,1

%A _N. J. A. Sloane_, Feb 24 2023