A291463 -id:A291463 - cp's OEIS frontend

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

Enoch Haga

			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.

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

Cf. A000040, A048851, A006094, A291463.

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

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 *)

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

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

A064009 a(n) = Sum_{k=1..n} d(k)*prime(k), where d(k) = A001223.

Original entry on oeis.org

2, 8, 18, 46, 68, 120, 154, 230, 368, 426, 612, 760, 842, 1014, 1296, 1614, 1732, 2098, 2366, 2508, 2946, 3262, 3760, 4472, 4860, 5062, 5474, 5688, 6124, 7706, 8214, 9000, 9274, 10664, 10962, 11868, 12810, 13462, 14464, 15502, 15860, 17670, 18052

Offset: 1

Jason Earls, Sep 06 2001

nonn

			a(4) = 2*(3-2) + 3*(5-3) + 5*(7-5) + 7*(11-7) = 46.

Harry J. Smith, Table of n, a(n) for n=1..1000

Cf. A001223.

Partial sums of A291463.

Accumulate@ Array[# (NextPrime@ # - #) &@ Prime@ # &, {43}] (* Michael De Vlieger, May 05 2016 *)

d(n) = prime(n+1)-prime(n); j=[]; for(n=1,150,j=concat(j,sum(k=1,n, prime(k)*d(k)))); j

d(n)= { prime(n + 1) - prime(n) }
{ a=0; for (n=1, 1000, write("b064009.txt", n, " ", a+=prime(n)*d(n)) ) } \\ Harry J. Smith, Sep 05 2009

p=primes_first_n(44); d=differences(p); v=[]; a=0
for k in range(len(d)):
   a+=p[k]*d[k]; v.append(a)
print(v)  # Giuseppe Coppoletta, May 05 2016

cp's OEIS Frontend

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

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