Lei Zhou - cp's OEIS frontend

A379985 Numbers k such that k^2 is of the form b^2 + (4c)^2 where bc is squarefree.

5, 13, 17, 25, 29, 37, 61, 65, 85, 109, 137, 145, 149, 157, 169, 173, 193, 197, 205, 221, 229, 241, 265, 269, 293, 305, 325, 365, 377, 401, 409, 421, 433, 445, 485, 505, 533, 541, 557, 565, 569, 629, 673, 685, 689, 701, 709, 725, 761, 773, 797

Offset: 1

Lei Zhou, Jan 07 2025

nonn

It is known that the sum of squares of two odd numbers cannot be a square number, and when the sum of square of two numbers is the square of an odd number, the even one among the two numbers has to be multiple of 4. Thus the Mathematica program will not miss any entries.
a(n) == 1 (mod 4).
Numbers 4x^2 + y^2 where x, y are coprime numbers such that y is odd and x, y, 2x+y, 2x-y are squarefree. - Yifan Xie, Jan 09 2025, corrected by Robert Israel, Feb 03 2025

			5 is a term since 5^2 = 3^2 + (4*1)^2 and 3*1 is squarefree.
149 is a term since 149^2 = 51^2 + (4*35)^2 and 51*35 = 3*5*7*17 is squarefree.

Robert Israel, Table of n, a(n) for n = 1..10000

Subsequence of A009003.

Cf. A005117, A084645, A009000, A057100.

N:= 1000: # for terms <= N
Res:= {}:
for x from 1 while 4*x^2 < N do
  if not numtheory:-issqrfree(x) then next fi;
  for y from 1 by 2  while 4*x^2 + y^2 <= N do
    if igcd(x,y) = 1 and andmap(numtheory:-issqrfree,[y,2*x+y,2*x-y]) then Res:= Res union  {4*x^2 + y^2} fi
od od:
sort(convert(Res,list)); # Robert Israel, Feb 03 2025

a = {}; Do[m = n^2; b = n; While[b = b - 2; b > 1, k = m - b^2; If[c = Sqrt[k]/4; IntegerQ[c] && SquareFreeQ[b*c], AppendTo[a, n]]], {n, 5, 800, 2}]; a

Edited by Robert Israel, Feb 03 2025

A379260 Index of first appearance of n in sequence A379049.

Original entry on oeis.org

0, 1, 3, 2, 9, 4, 6, 26, 24, 5, 18, 7, 78, 28, 12, 11, 54, 19, 216, 71, 15, 29, 162, 53, 21, 73, 231, 16, 486, 13, 51, 217, 84, 83, 36, 14, 33, 647, 57, 32, 4374, 31, 237, 649, 45, 22, 207, 236, 165, 1945, 693, 50, 2151, 212, 90, 46, 87, 160, 39366, 86, 63

Offset: 2

Lei Zhou, Dec 19 2024

a(n) is the smallest integer that makes A379049(a(n)) = n.

Conjecture: a(n) is defined for all integer n > 1.

			For n = 2, A379049(0) = 1 + 1 = 2.  Thus a(2) = 0;
For n = 3, A379049(1) = 2 + 1 = 3, since 1's balanced ternary representation is 1.  Thus a(3) = 1;
For n = 4, A379049(3) = 3 + 1 = 4, since 3's balanced ternary representation is 10.  Thus a(4) = 3;
...
For n = 60, A379049(39366) = 31 + 29 = 60, since 39366's balanced ternary representation is 1T000000000, where the 11's digit is 1 represents the 11's prime 31 in the term before the plus sign, and the 10's digit is T representing the 10's prime 29 in the term after the plus sign. And evaluation of A379049 found no number i smaller than 39366 can make A379049(i) = 60.  Thus a(60) = 39366.

Cf. A379049, A345128, A005812, A134022, A134023, A140267.

BTDigits[m_Integer, g_] :=  Module[{n = m, d, sign, t = g}, If[n != 0, If[n > 0, sign = 1, sign = -1; n = -n]; d = Ceiling[Log[3, n]]; If[3^d - n <= ((3^d - 1)/2), d++];    While[Length[t] < d, PrependTo[t, 0]]; t[[Length[t] + 1 - d]] = sign; t = BTDigits[sign*(n - 3^(d - 1)), t]]; t];
goal = 62; res = {}; ct = 1;
Do[AppendTo[res, 0], {i, 2, goal}]; i = -1; While[ct < goal, i++; BT = BTDigits[i, {0}]; BTl = Length[BT]; f = 1; b = 1;  Do[If[BT[[j]] == 1, f = f*Prime[BTl - j + 1]];   If[BT[[j]] == -1, b = b*Prime[BTl - j + 1]], {j, 1, BTl}]; d = f + b; If[(d <= goal) && (res[[d - 1]] == 0), res[[d - 1]] = i; ct++]];
res

A379049 a(n) = prime(i)dp(n,i) + prime(i)dn(n,i) where dp(n,i) = 1 when the i-th trit of n is 1, dn(n,i) = 1 when the i-th trit of n is T, and dp(n,i) = dn(n,i) = 0 when the i-th trit of n is 0.

Original entry on oeis.org

2, 3, 5, 4, 7, 11, 8, 13, 7, 6, 11, 17, 16, 31, 37, 22, 29, 17, 12, 19, 31, 26, 47, 13, 10, 17, 9, 8, 15, 23, 22, 43, 41, 38, 73, 37, 36, 71, 107, 106, 211, 221, 116, 127, 81, 46, 57, 103, 68, 101, 53, 32, 43, 25, 18, 29, 47, 40, 73, 97, 76, 131, 69, 62, 117

Offset: 0

Lei Zhou, Dec 14 2024

The Balanced Ternary presentation of a number is a series of 1, 0, and T, where T represent -1. For example, 35 = 110T = 1 * 3^3 + 1* 3^2 + 0 * 3 - 1 = 27 + 9 + 0 - 1.

Conjecture: All positive integers greater than 1 appear in this sequence at least once.

			When n = 0, its BT presentation is 0, thus a(0) = 1 + 1 = 2;
When n = 1, its BT presentation is 1, the first prime is 2, thus a(1) = 2 + 1 = 3;
...
When n = 14, its BT presentation is 1TTT, thus prime 7 appears before the plus sign and primes 5, 3, and 2 appear in the term after the plus sign, a(14) = 7 + 5*3*2 = 37;
...
By the same rule, when n = 64, its BT presentation is 1T101, thus prime 11, 5, 2 appear before the plus sign and prime 7 appears in the term after the plus sign, a(64) = 11*5*2 + 7 = 117.

John Tyler Rascoe, Table of n, a(n) for n = 0..10000

Cf. A005812, A134022, A134023, A140267.

BTDigits[m_Integer, g_]:= Module[{n = m, d, sign, t = g}, If[n != 0, If[n > 0, sign = 1, sign = -1; n = -n]; d = Ceiling[Log[3, n]]; If[3^d - n <= ((3^d - 1)/2), d++]; While[Length[t] < d, PrependTo[t, 0]]; t[[Length[t] + 1 - d]] = sign; t = BTDigits[sign*(n - 3^(d - 1)), t]]; t];
res = {}; Do[BT = BTDigits[i, {0}]; BTl = Length[BT]; f = 1; b = 1; Do[If[BT[[j]] == 1, f = f*Prime[BTl - j + 1]]; If[BT[[j]] == -1, b = b*Prime[BTl - j + 1]], {j, 1, BTl}];  d = f + b; AppendTo[res, d], {i, 0, 64}]; res

from sympy import prime
def A140267(n): # see A140267
    return
def A379049(n):
    x,y,z = 1,1,str(A140267(n))[::-1]
    for i in range(len(z)):
        if z[i] == "1":
            x *= prime(i+1)
        if z[i] == "2":
            y *= prime(i+1)
    return x+y # John Tyler Rascoe, Feb 27 2025

A377014 a(n) is the number of primes p such that p - 6, p + 6 and 2*n - p are also primes.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 1, 2, 2, 2, 2, 3, 2, 3, 4, 1, 3, 3, 0, 4, 4, 2, 2, 3, 3, 3, 6, 3, 4, 6, 0, 5, 5, 1, 6, 4, 3, 5, 6, 4, 3, 9, 3, 2, 8, 2, 4, 7, 2, 4, 3, 3, 5, 5, 6, 4, 9, 4, 4, 11, 2, 5, 10, 1, 4, 4, 4, 4, 4, 5, 2, 7, 4, 4, 9, 2, 5, 6, 0, 6, 7, 5, 3, 6, 5, 1, 10, 7, 4, 9, 2, 5, 9, 2, 6, 5, 4, 5, 4, 4

Offset: 1

Lei Zhou, Oct 12 2024

Conjecture: a(n) = 0 only when n = 1, 2, 3, 4, 5, 6, 19, 31, 331, 499.

			a(7) = 1 since only when p = 11 are p - 6, p + 6 and 2n - p all prime.
a(12) = 3 from the cases when p is 11, 13 or 17:
  when p = 11, {p - 6, p + 6, 2n - p} = {5, 17, 13} are all prime;
  when p = 13, {p - 6, p + 6, 2n - p} = {7, 13, 19, 11} are all prime;
  when p = 17, {p - 6, p + 6, 2n - p} = {11, 17, 23, 7} are all prime.
a(19) = 0 since 2n = 38 = 7 + 31 = 19 + 19 = 31 + 7, and none of p = 7, 19, 31 can make p - 6 and p + 6 both prime.

Cf. A006489, A045917.

f:= proc(n) local i;
  nops(select(p -> andmap(isprime,[p,p-6,p+6, 2*n-p]), [seq(i,i=3..2*n,2)]))
end proc:
map(f, [$1..100]); # Robert Israel, Oct 13 2024

m = 200; ps = {}; p = 7; While[p = NextPrime[p]; If[PrimeQ[p - 6] && PrimeQ[p + 6], AppendTo[ps, p]]; p < 2*m]; a = {}; Do[ct = 0; k = 0; While[k++; ps[[k]] < n, q = n - ps[[k]]; If[PrimeQ[q], ct++]]; AppendTo[a, ct]; If[ct == 0, AppendTo[b, n]], {n, 2, m, 2}]; a

A376917 Starting from Goldbach decomposition of 10 = p + q = 5 + 5, 12 = 7 + 5, and 14 = 7 + 7, a(n) is the first number in A001057 such that if 2n - 6 = p + q, 2n = p' + q', where p' = p + 6 * a(n) and q' = 2n - q' are both primes.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 1, -1, 1, 1, 0, 1, -1, 0, 1, 0, 1, -1, 0, 2, 0, 0, 0, 0, 1, 1, 0, -1, 1, 1, 1, -2, -1, -1, 0, 1, 0, 0, -1, 1, 0, 0, -1, 1, 1, 0, 1, -1, 0, -2, 0, 1, 0, 0, 2, 0, 1, 0, 1, 2, 1, 1, 0, 1, -2, 1, -2, 1, 1, 0, 1, 1, 1, -1, 2, 1, 1

Offset: 8

Lei Zhou, Oct 09 2024

By definition, this sequence starts from n = 8.

Hypothesis: a(n) is defined for all n >= 8 and for all n >=8, the corresponding Goldbach decomposition 2n = p + q has positive primes p and q.

			When n = 8, 2n = 16.  2n - 6 = 10. 10 = p + q = 5 + 5 (by definition). a(8) = 0, p' = p + a(8) = 5, q' = 2n - p' = 16 - 5 = 11. P' and q' are both primes.
...
When n = 10, 2n = 20. 2n - 6 = 14. 14 = p + q = 7 + 7 (by definition). a(10) = 0, p' = p + a(10) = 7, q' = 2n - p' = 20 - 7 = 13. P' and q' are both primes.
...
When n = 13, 2n = 26. 2n - 6 = 20. 20 = p + q = 7 + 13 (per above evaluation). a(13) = 0, p' = p + a(13) = 7, q' = 2n - p' = 26 - 7 = 19. P' and q' are both primes.
When n = 16, 2n = 32. 2n - 6 = 26. 26 = p + q = 7 + 19 (per above evaluation). a(16) = 1, p' = p + a(16) = 13, q' = 2n - p' = 32 - 13 = 19. P' and q' are both primes. It is tested when a(16) is 0, q' = 25 is not a prime, thus a(16) = 1 is the first number in A001057 that makes both p' and q' primes.

Ref. A001057; Cf. A045917

a = {}; p = {5, 7, 7}; Do[Do[n = 6*k - 4 + 2*j; i = 0; While[i++; m = 1/4 + (i - 1/2)*(-1)^i/2; pr = p[[j]] + 6*m; q = n - pr; ! (PrimeQ[pr] && PrimeQ[q])]; p[[j]] = pr; AppendTo[a, m], {j, 1, 3}], {k, 3, 30}]; Print[a]

A376510 a(n) is the number of pairs of primes p+q=2*(n+4) with 5 <= p <= n such that either p+6 or q+6 is also prime.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 2, 3, 2, 2, 3, 1, 3, 4, 1, 2, 4, 2, 3, 4, 3, 3, 5, 2, 3, 6, 1, 4, 6, 2, 4, 5, 4, 4, 6, 4, 4, 8, 3, 3, 8, 3, 5, 7, 2, 4, 7, 4, 5, 6, 5, 6, 9, 5, 4, 12, 3, 5, 10, 2, 5, 7, 5, 5, 6, 6, 5, 11, 5, 4, 11, 2, 7, 8, 3, 6, 10, 5, 4, 9, 7, 5, 11, 6

Offset: 1

Lei Zhou, Sep 25 2024

It is hypothesized that all terms of this sequence are positive integers.

If the above hypothesis is true, the Goldbach Hypothesis is true, since for every even number 2n, if there is a Goldbach decomposition p+q=2n meets the condition of this sequence, p+q+6=2n+6 forms at least one Goldbach decomposition of 2n+6.

			For n=1, 2*(n+4)=10, 10=5+5, and 5+6=11 is a prime. Thus a(1)=1;
For n=2, 2*(n+4)=12, 12=5+7, and 5+6=11 is a prime. Thus a(2)=1;
...
For n=14, 2*(n+4)=36, 36=5+31 (5+6=11); 7+29 (7+6=13); 13+23 (13+6=19); 17+19 (17+6=23), four cases found.  Thus a(14)=4.

Cf. A240712, A171611, A254041.

res = {}; Do[n[2] = i*6; n[1] = n[2] - 2; n[3] = n[2] + 2;
 Do[c[j] = 0; p[j] = NextPrime[n[j]/2 - 1];
  While[q[j] = n[j] - p[j];
   If[PrimeQ[q[j]] && q[j] > 3,
    If[PrimeQ[p[j] + 6] || PrimeQ[q[j] + 6], c[j]++]];
   p[j] < n[j] - 5, p[j] = NextPrime[p[j]]], {j, 1, 3}];
 AppendTo[res, c[1]]; AppendTo[res, c[2]];
 AppendTo[res, c[3]], {i, 2, 29}]; Print[res]

A337124 a(n) is the number of primes p in the n-digit "signed nonadjacent form" such that p has three nonzero digits.

Original entry on oeis.org

0, 0, 0, 0, 3, 4, 7, 4, 8, 8, 12, 7, 11, 6, 11, 9, 13, 9, 18, 10, 21, 7, 9, 11, 16, 4, 8, 9, 7, 12, 18, 12, 14, 11, 10, 9, 18, 7, 12, 10, 18, 12, 22, 5, 11, 13, 16, 13, 22, 8, 9, 16, 13, 9, 13, 14, 10, 11, 10, 10, 20, 14, 9, 10, 13, 8, 22, 10, 10, 10, 12, 13

Offset: 1

Lei Zhou, Aug 17 2020

Sign nonadjacent form notation is defined by the publications listed in the reference.

We use abbreviation SNF for "signed nonadjacent form" notation.

			It needs at least 5 digits to have three or more nonzero digits in SNF notation.  So a(1)=a(2)=a(3)=a(4)=0.
In 5-digit SNF numbers, 10T0T = 11 base 10, 10T01 = 13, and 10101 = 19 are primes with three nonzero digits in SNF notation.  So a(5)=3.  Another prime with 5 SNF digits, 10001 = 17 has only 2 SNF digits, so is excluded.

Joerg Arndt, Matters Computational - Ideas, Algorithms, Source Code, 2011, Springer, pp. 61-62.

H. Prodinger, On binary representations of integers with digits -1,0,1, Integers 0 (2000), #A08.

Cf. A334913, A337123.

Table[s1=2^(n-1);ct=0; If[n>=5, Do[s2=2^i; If[PrimeQ[s1+s2+1],ct++]; If[PrimeQ[s1+s2-1],ct++]; If[PrimeQ[s1-s2+1],ct++]; If
[PrimeQ[s1-s2-1],ct++], {i,2,n-3}]]; ct, {n,1,73}]

A337123 a(n) is the number of primes p in the n-digit "signed nonadjacent form" such that p has 3 or fewer nonzero digits.

Original entry on oeis.org

0, 1, 2, 1, 4, 5, 7, 5, 9, 8, 12, 7, 11, 7, 11, 9, 14, 10, 18, 11, 21, 7, 9, 11, 16, 4, 8, 9, 7, 12, 18, 13, 14, 11, 10, 9, 18, 7, 12, 10, 18, 12, 22, 5, 11, 13, 16, 13, 22, 8, 9, 16, 13, 9, 13, 14, 10, 11, 10, 10, 20, 15, 9, 10, 13, 8, 22, 10, 10, 10, 12, 13

Offset: 1

Lei Zhou, Aug 17 2020

Sign nonadjacent form notation is defined by the publications listed in the reference.
We use abbreviation SNF for "signed nonadjacent form" notation.
This is an expansion of A337124 to include 2 and primes in the form of 2^k+1 and 2^k-1.

			There is only one number in single-digit SNF notation, which is 1 and 1 is not a prime.  So a(1)=0;
There is only one number in the two-digit SNF notation, which is 10 = 2 base 10 and it is a prime with one nonzero digit.  So a(2)=1;
There are three numbers in three digits SNF notation: 10T = 3 base 10, 100 = 4 base 10, and 101 = 5 base 10.  There are two prime numbers among 3, 4, and 5 and both of them have two nonzero digits. So a(3)=2;
...
For seven-digit SNF numbers, 10T0T0T = 43 base 10 has 4 nonzero digits (excluded); 10T000T = 47 base 10 has 3 nonzero digits (included). Thereafter 10T0101 = 53: 4 digits, excluded; 1000T0T = 59: 3 digits, included; 1000T01 = 61: 3 digits, included; 100010T = 67: 3 digits, included; 100100T = 71: 3 digits, included; 1001001 = 73, 3 digits, included; 101000T = 79: 3 digits, included; 101010T = 83, 4 digits, excluded.  In total, 7 numbers fit the definition.  So a(7)=7.

Joerg Arndt, Matters Computational - Ideas, Algorithms, Source Code, 2011, Springer, pp. 61-62.

H. Prodinger, On binary representations of integers with digits -1,0,1, Integers 0 (2000), #A08.

Cf. A334913, A337124.

Table[s1=2^(n-1);ct=0; If[n>1, If[PrimeQ[s1+1],ct++]; If[PrimeQ[s1-1],ct++]; If[n>=5, Do[s2=2^i; If[PrimeQ[s1+s2+1],ct++]; If[PrimeQ[s1+s2-1],ct++]; If[PrimeQ[s1-s2+1],ct++]; If[PrimeQ[s1-s2-1],ct++], {i,2,n-3}]]]; ct, {n,1,72}]

A337121 a(n) is the number of ways the n-th prime number prime(n) can be represented as sum of two smaller odd prime numbers p1, p2 with prime(n) > p1 > (p2 minus the maximum odd prime factor of (p1-p2)).

Original entry on oeis.org

1, 1, 2, 3, 2, 2, 4, 3, 3, 5, 3, 6, 4, 4, 4, 7, 4, 4, 5, 6, 6, 6, 9, 7, 8, 8, 7, 7, 6, 11, 4, 11, 9, 7, 8, 9, 7, 13, 12, 6, 10, 15, 10, 9, 7, 13, 13, 11, 13, 10, 15, 10, 13, 14, 11, 13, 13, 12, 14, 17, 13, 13, 19, 9, 14, 19, 12, 8, 14, 22, 17, 14, 13, 16, 9, 15

Offset: 4

Lei Zhou, Aug 17 2020

nonn

This sequence counts the cases such that prime(n) = p1 + p2 - MaxOddPrimeFactor(p1-p2), where MaxOddPrimeFactor(m) is defined as the maximum odd prime factor of the positive integer m. If there is no odd prime factor of m, MaxOddPrimeFactor(m) is defined as 1.
Conjecture: a(n) > 0 when n >= 4.
Some nonprime odd numbers, like 27, cannot be partitioned into the form of p1 + p2 - MaxOddPrimeFactor(p1-p2).

			When n=4, prime(4)=7, MaxOddPrimeFactor(5-3)=1, 7=5+3-1. This is the only case, so a(4)=1.
When n=5, prime(5)=11, MaxOddPrimeFactor(7-5)=1, 11=7+5-1. This is the only case, so a(5)=1.
When n=6, prime(6)=13, MaxOddPrimeFactor(11-3)=1, 13=11+3-1; and MaxOddPrimeFactor(11-5)=3, 13=11+5-3. Two cases found, so a(6)=2.

MaxOddPrimeFactor[m_] :=
Module[{factors, l, res}, factors = FactorInteger[m];
  l = Length[factors]; res = factors[[l, 1]]; If[res == 2, res = 1];
  res]
Table[p = Prime[n]; p1 = NextPrime[p/2, -1]; ct = 0;
While[p1 = NextPrime[p1]; p1 < p, p2 = NextPrime[p - p1, -1];
  While[p2 = NextPrime[p2]; p2 < p1,
   If[p == (p1 + p2 - MaxOddPrimeFactor[p1 - p2]), ct++]]]; ct, {n, 4,
   79}]

A334913 a(n) is the sum of digits of n in signed binary nonadjacent form.

Original entry on oeis.org

0, 1, 1, 0, 1, 2, 0, 0, 1, 2, 2, -1, 0, 1, 0, 0, 1, 2, 2, 1, 2, 3, -1, -1, 0, 1, 1, -1, 0, 1, 0, 0, 1, 2, 2, 1, 2, 3, 1, 1, 2, 3, 3, -2, -1, 0, -1, -1, 0, 1, 1, 0, 1, 2, -1, -1, 0, 1, 1, -1, 0, 1, 0, 0, 1, 2, 2, 1, 2, 3, 1, 1, 2, 3, 3, 0, 1, 2, 1, 1, 2, 3, 3, 2

Offset: 0

Lei Zhou, May 16 2020

Joerg Arndt, Matters Computational - Ideas, Algorithms, Source Code, 2011, Springer, pp. 61-62.
Helmut Prodinger, On binary representations of integers with digits -1,0,1, Integers 0 (2000), #A08.

Cf. A000120, A001045, A007302, A184615, A184616.

BBN[a_] := Module[{n = a, b}, b = IntegerDigits[n, 2]; b = Prepend[b, 0];
  l = Length[b];
  Do[If[b[[i]] == 2, b[[i]] = 0; b[[i - 1]]++,
    If[b[[i]] == 1,
     If[b[[i + 1]] == 1, b[[i - 1]]++; b[[i]] = 0;
      b[[i + 1]] = -1]]], {i, l - 1, 2, -1}];
  If[b[[1]] == 0, b = Delete[b, 1]]; b]
Table[a = BBN[i]; sod = 0; l = Length[a];
Do[sod = sod + a[[j]], {j, 1, l}]; sod, {i, 0, 83}]

bin2naf(x)=
{ /* Compute (nonadjacent) signed binary representation of x: */
    local(xh, x3, c, np, nm);
    xh = x >> 1;
    x3 = x + xh;
    c = bitxor(xh, x3);
    np = bitand(x3, c);  /* bits == +1 */
    nm = bitand(xh, c);  /* bits == -1 */
    return([np, nm]);  /* np-nm==x */
}
a(n) = my(b=bin2naf(n)); return(hammingweight(b[1])-hammingweight(b[2]));
vector(99,n,a(n-1)) \\ Joerg Arndt, Jun 13 2020

a(n) = hammingweight(A184615(n)) - hammingweight(A184616(n)). - Joerg Arndt, Jun 13 2020

cp's OEIS Frontend

User: Lei Zhou

A379985 Numbers k such that k^2 is of the form b^2 + (4*c)^2 where b*c is squarefree.

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A379260 Index of first appearance of n in sequence A379049.

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A379049 a(n) = prime(i)*dp(n,i) + prime(i)*dn(n,i) where dp(n,i) = 1 when the i-th trit of n is 1, dn(n,i) = 1 when the i-th trit of n is T, and dp(n,i) = dn(n,i) = 0 when the i-th trit of n is 0.

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A377014 a(n) is the number of primes p such that p - 6, p + 6 and 2*n - p are also primes.

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A376917 Starting from Goldbach decomposition of 10 = p + q = 5 + 5, 12 = 7 + 5, and 14 = 7 + 7, a(n) is the first number in A001057 such that if 2n - 6 = p + q, 2n = p' + q', where p' = p + 6 * a(n) and q' = 2n - q' are both primes.

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A376510 a(n) is the number of pairs of primes p+q=2*(n+4) with 5 <= p <= n such that either p+6 or q+6 is also prime.

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A337124 a(n) is the number of primes p in the n-digit "signed nonadjacent form" such that p has three nonzero digits.

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A337123 a(n) is the number of primes p in the n-digit "signed nonadjacent form" such that p has 3 or fewer nonzero digits.

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Examples

References

A379985 Numbers k such that k^2 is of the form b^2 + (4c)^2 where bc is squarefree.

A379049 a(n) = prime(i)dp(n,i) + prime(i)dn(n,i) where dp(n,i) = 1 when the i-th trit of n is 1, dn(n,i) = 1 when the i-th trit of n is T, and dp(n,i) = dn(n,i) = 0 when the i-th trit of n is 0.