A048852 Difference between b^2 (in c^2=a^2+b^2) and product of successive prime pairs.

0, 3, 10, 14, 44, 26, 68, 38, 92, 174, 62, 222, 164, 86, 188, 318, 354, 122, 402, 284, 146, 474, 332, 534, 776, 404, 206, 428, 218, 452, 1778, 524, 822, 278, 1490, 302, 942, 978, 668, 1038, 1074, 362, 1910, 386, 788, 398, 2532, 2676, 908, 458, 932, 1434, 482

Offset: 0

Author

Enoch Haga

Keywords

• easy
• nonn

Examples

			a(3)=10. Product of 3rd prime pair 3*5=15 (after 2*2=4 and 2*3=6). b^2=25 (in c^2=a^2+b^2) where c^2=34 and a^2=9. Then 25-15=10.
		

Links

• G. C. Greubel, Table of n, a(n) for n = 0..10000

Crossrefs

Cf. A000040, A048851, A006094, A291463.

Programs

Magma

[0] cat [NthPrime(n+1)*(NthPrime(n+1)-NthPrime(n)): n in [1..60]]; // G. C. Greubel, Feb 22 2024

Mathematica

With[{P=Prime}, Table[If[n==0, 0, P[n+1]*(P[n+1]-P[n])], {n,0,60}]] (* G. C. Greubel, Feb 22 2024 *)

SageMath

p=nth_prime; [0]+[p(n+1)*(p(n+1)-p(n)) for n in range(1,61)] # G. C. Greubel, Feb 22 2024

Formula

Find b^2 in Pythagorean formula c^2=a^2+b^2. Subtract product of successive prime pair at same a(n) beginning at 2*2.

For n>0, a(n) = A000040(n+1)^2 - A000040(n) * A000040(n+1). - Mamuka Jibladze, Mar 24 2017