A048851 -id:A048851 - cp's OEIS frontend

A069484 a(n) = prime(n+1)^2 + prime(n)^2.

13, 34, 74, 170, 290, 458, 650, 890, 1370, 1802, 2330, 3050, 3530, 4058, 5018, 6290, 7202, 8210, 9530, 10370, 11570, 13130, 14810, 17330, 19610, 20810, 22058, 23330, 24650, 28898, 33290, 35930, 38090, 41522, 45002

Offset: 1

Reinhard Zumkeller, Mar 29 2002

Together with A069482(n) and A069486(n) a Pythagorean triangle is formed with area = A069487(n).

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
Janyarak Tongsomporn, Saeree Wananiyaku, and Jörn Steuding, Sums of consecutive prime squares, Integers (2022) Vol. 22, #A9.

Cf. A069485, A075892, A069482, A069486.

Cf. A001248, A000040, A048851.

seq(ithprime(n)^2+ithprime(n+1)^2, n = 1 .. 100); # Stefano Spezia, Dec 21 2018

Table[Prime[n]^2+Prime[n+1]^2,{n,5!}] (* Vladimir Joseph Stephan Orlovsky, Apr 12 2010 *)
Total[#^2]&/@Partition[Prime[Range[50]],2,1] (* Harvey P. Dale, May 26 2012 *)

v=primes(101);vector(#v-1,i,v[i]^2+v[i+1]^2) \\ Charles R Greathouse IV, Aug 21 2011

from sympy import prime
for n in range(1,101): print(n, prime(n)**2+prime(n+1)**2) # Stefano Spezia, Dec 21 2018

a(n) = A001248(n+1) + A001248(n) = A000040(n+1)^2 + A000040(n)^2.
a(n) = A048851(n+1).
a(n) = 2 * A075892(n) for n > 1.

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

Original entry on oeis.org

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

A048871 Length of hypotenuse squared in right triangle formed by a palindromic spiral plotted in Cartesian coordinates.

Original entry on oeis.org

2, 5, 13, 25, 41, 61, 85, 113, 145, 202, 605, 1573, 3025, 4961, 7381, 10285, 13673, 17545, 20002, 22522, 26962, 31802, 37042, 42682, 48722, 55162, 62002, 69242, 77285, 85748, 94228, 103108, 112388, 122068, 132148, 142628, 153508, 164788

Offset: 1

Enoch Haga

			To begin palindromic spiral, plot (1, 0), (0, 1). Hypotenuse is c^2=a^2+b^2, or 2=1+1. a(1) = 2.
a(2)=5 because c^2=a^2+b^2 and 5=1+4.

H. E. Huntley, The Divine Proportion, A Study in Mathematical Beauty. New York: Dover, 1970. See Chapter 13, Spira Mirabilis, especially Fig. 13-5, page 173.

Harvey P. Dale, Table of n, a(n) for n = 1..1000
Patrick De Geest, World!Of Numbers

Cf. A002113.

An analog of the prime spiral of A048851.

Join[{2},Total[#^2]&/@Partition[Select[Range[300],PalindromeQ],2,1]] (* Harvey P. Dale, Mar 18 2024 *)

a(n) = A002113(n)^2 + A002113(n-1)^2, n > 1.

cp's OEIS Frontend

A069484 a(n) = prime(n+1)^2 + prime(n)^2.

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

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A048871 Length of hypotenuse squared in right triangle formed by a palindromic spiral plotted in Cartesian coordinates.

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Mathematica

Formula