Derek Orr - cp's OEIS frontend

A339928 Numbers k such that the removal of all terminating even digits from k! leaves a prime.

6, 7, 9, 10, 43, 138, 1068

Offset: 1

Derek Orr, Dec 23 2020

a(8) > 1500.
If only the terminating zeros are removed, 2 is the only number whose factorial becomes prime.
If one also removes 5s at the end, 7 is no longer in the sequence and no numbers below 1500 are added to the sequence.
a(8) > 20000. - Michael S. Branicky, Jul 05 2024

			43! = 60415263063373835637355132068513997507264512000000000. After removing all even digits at the end, we are left with 6041526306337383563735513206851399750726451, which is prime. So 43 is a term of this sequence.
27! = 10888869450418352160768000000. After removing all even digits at the end, we are left with 108888694504183521607, which is not prime. So 27 is not a term of this sequence.

Cf. A000142.

for(n=1,1500,k=n!;while(!(k%2),k\=10;if(k==0,break));if(isprime(k),print1(n,", ")))

from sympy import factorial, isprime
def ok(n):
    fn = factorial(n)
    while fn > 0 and fn%2 == 0: fn //= 10
    return fn > 0 and isprime(fn)
print(list(filter(ok, range(200)))) # Michael S. Branicky, Jun 07 2021

A338941 a(1)=1. For n >= 2, let S be the sum of all prime digits in a(1), a(2), ... a(n-1) and let C be the next nonprime number not already in the sequence. If S is a prime less than C and is not already a term of the sequence, a(n) = S. Otherwise, a(n) = C.

Original entry on oeis.org

1, 4, 6, 8, 9, 10, 12, 2, 14, 15, 16, 18, 20, 11, 21, 13, 22, 24, 25, 26, 27, 28, 30, 32, 33, 34, 35, 36, 38, 39, 40, 42, 44, 45, 46, 48, 49, 50, 51, 52, 54, 55, 56, 57, 58, 60, 62, 63, 64, 65, 66, 68, 69, 70, 72, 74, 75, 76, 77, 78, 80, 81, 82, 84, 85, 86, 87, 88, 90

Offset: 1

Eric Angelini and Derek Orr, Nov 17 2020

Similar to A338924, however this sequence does not account for the prime digits of a(n) itself.

Each prime term is the sum of all prime digits of each previous term.

			a(16) = 13 because the sum of the prime digits from the previous terms is 2+2+5+2+2 = 13 (a prime) and 13 is less than the next nonprime (22).
a(17) = 22 because the sum of the prime digits from the previous terms is 2+2+5+2+2+3 = 16 (a nonprime), so a(17) is the next nonprime in the sequence.
a(18) = 24 because the sum of the prime digits from the previous terms is 2+2+5+2+2+3+2+2 = 20 (a nonprime).
a(16) = 25 because the sum of the prime digits from the previous terms is 2+2+5+2+2+3+2+2+2 = 22 (a nonprime).
a(17) = 26 because the sum of the prime digits from the previous terms is 2+2+5+2+2+3+2+2+2+2+5 = 29 (a prime) but it is not less than the next nonprime (which is 26).

Cf. A338924.

a(n)=my(v=[1], S=0, k=1, C=4, m); while(k

A338938 a(1)=0. For n >= 2, let S be the sum of all nonprime digits in a(1), a(2), ... a(n-1) and let P be the next prime not already in the sequence. If S is a nonprime number less than P and not already in the sequence, a(n) = S. Otherwise, a(n) = P.

Original entry on oeis.org

0, 2, 3, 5, 7, 11, 13, 17, 4, 8, 16, 19, 23, 29, 31, 37, 41, 43, 47, 53, 56, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257

Offset: 1

Eric Angelini and Derek Orr, Nov 16 2020

Similar to A338925, however this sequence does not include the nonprime digits of a(n) itself.

Each nonprime term is the sum of all nonprime digits of each previous term.

			a(9) = 4 since the sum of the nonprime digits of the previous terms is 1+1+1+1 =  4 and 4 is less than the next prime, 19.
a(10) = 8 since the sum of nonprime digits of the previous terms is 1+1+1+1+4 = 8 and 8 is less than the next prime, 19.
a(11) = 16 since the sum of the nonprime digits of the previous terms is 1+1+1+1+4+8 = 16 and 16 is less than the next prime, 19.
Now, the sum of the nonprime digits of the previous terms is 1+1+1+1+4+8+1+6 = 23 (a prime number). So a(12) is the next prime number in that hasn't appeared, meaning a(12) = 19.

Cf. A338925.

my(v=[0], w=[0], n=1, p=1, m, c); while(n<125, q=vecsum(w);m=[];p=nextprime(p);c=0; for(k=1,#digits(q),if(!isprime(digits(q)[k]),m=concat(m,digits(q)[k])));if(!isprime(q)&&(q

A338921 a(0)=1, a(n) for n >= 1 is the number of distinct sums of two elements in [a(0), ..., a(n-1)], chosen without replacement.

Original entry on oeis.org

1, 0, 1, 2, 3, 5, 8, 12, 17, 22, 28, 35, 43, 52, 60, 69, 77, 86, 92, 103, 112, 123, 137, 151, 168, 180, 194, 204, 224, 245, 261, 280, 301, 318, 335, 352, 369, 387, 413, 433, 459, 482, 507, 528, 552, 586, 614, 638, 669, 701, 733, 761, 791, 824, 855, 885, 917, 952, 985, 1020

Offset: 0

Derek Orr, Nov 15 2020

nonn

a(n) <= A000217(n)-n for n >= 1.
Without replacement means a(i)+a(i) is not included. However, if a(i)=a(j), a(i)+a(j) still counts because they have two different indices. If you include a(i)+a(i), the sequence becomes A000012 (all ones).
If you include the distinct sums between 3 elements and more, you arrive at the sequence 1, 0, followed by A000079 (2^n).
Same rule as in A247184, but with a(0)=1.

			a(1) gives the number of distinct sums between two elements of [1]. There aren't two elements so a(1)=0.
a(2) gives the number of distinct sums between two elements of [1,0]. The only sum are 1+0, so a(2) = 1.
a(3) gives the number of distinct sums between two elements of [1,0,1]. The two sums are 1+0 and 1+1 so a(3)=2.

Cf. A000217, A000012, A247184, A247185.

s:= proc(n) option remember; `if`(n=0, {},
      {s(n-1)[], seq(a(i)+a(n), i=0..n-1)})
    end:
a:= proc(n) option remember;
      `if`(n=0, 1, nops(s(n-1)))
    end:
seq(a(n), n=0..60);  # Alois P. Heinz, Nov 16 2020

a[0] = 1; a[1] = 0;
a[n_Integer?Positive] := a[n] = Length[Union[Total[Subsets[Array[a, n, 0], {2}], {2}]]];
Array[a, 61, 0] (* Jan Mangaldan, Nov 23 2020 *)

my(v=[1], w=[], n=1); while(n<75, for(i=2, #v, w=concat(w,v[i-1]+v[#v])); w=vecsort(w,,8); v=concat(v, #w); n++); v

A290227 Numbers n such that A290223(n) = 11.

Original entry on oeis.org

8, 17, 26, 29, 35, 38, 47, 56, 65, 74, 83, 92, 149, 158, 167, 197

Offset: 1

Derek Orr, Jul 24 2017

This sequence is believed to be finite. If it exists, a(17) > 10^5.

			26 is in this sequence because 26 - (2+6)^2 = -38. Then -38 + (3+8)^2 = 83. Then 83 - (8+3)^2 = -38, and so on.

Cf. A007953, A003132, A290223.

a(n)=k=n; c=1; v=List(); listput(v, k); while(c, if(k>=0, k-=sumdigits(k)^2; c+=1; if(k==2||k==3||k==0||k==6||k==9, return(k)); if(vecsearch(Vec(v), k), return(sumdigits(abs(k)))); listput(v, k)); if(k<0, k+=sumdigits(-k)^2; c+=1; if(k==2||k==3||k==0||k==6||k==9, return(k)); if(vecsearch(Vec(v), k), return(sumdigits(abs(k)))); listput(v, k)); c+=1)
for(n=1,10^5,if(a(n)==11,print1(n,", ")))

A290226 Numbers n such that A290223(n) = 2.

Original entry on oeis.org

2, 23, 62, 77, 119, 194, 287, 398

Offset: 1

Derek Orr, Jul 24 2017

This sequence is believed to be finite. a(9) > 10^5, if it exists.

			62 is in this sequence because 62 - (6+2)^2 = -2. Then -2 + (2)^2 = 2.

Cf. A007953, A003132, A290223.

a(n)=k=n; c=1; v=List(); listput(v, k); while(c, if(k>=0, k-=sumdigits(k)^2; c+=1; if(k==2||k==3||k==0||k==6||k==9, return(k)); if(vecsearch(Vec(v), k), return(sumdigits(abs(k)))); listput(v, k)); if(k<0, k+=sumdigits(-k)^2; c+=1; if(k==2||k==3||k==0||k==6||k==9, return(k)); if(vecsearch(Vec(v), k), return(sumdigits(abs(k)))); listput(v, k)); c+=1)
for(n=1,10^5,if(a(n)==2,print1(n,", ")))

A290224 Numbers n such that A290223(n) = 0.

Original entry on oeis.org

1, 19, 81, 162, 181, 199, 243, 262, 324, 343, 405, 424, 486, 505, 567, 648, 685, 729, 766, 810, 847, 891, 910, 928, 972, 1053, 1072, 1134, 1153, 1215, 1234, 1296, 1315, 1377, 1458, 1495, 1539, 1576, 1620, 1657, 1701, 1720, 1738, 1782, 1801, 1819, 1863, 1944, 1981, 1999, 2025, 2044, 2106, 2125, 2187, 2206, 2268

Offset: 1

Derek Orr, Jul 24 2017

This sequence is believed to be infinite.

			181 is in this sequence because 181 - (1+8+1)^2 = 81. Then 81 - (8+1)^2 = 0.

Cf. A007953, A003132, A290223.

a(n)=k=n; c=1; v=List(); listput(v, k); while(c, if(k>=0, k-=sumdigits(k)^2; c+=1; if(k==2||k==3||k==0||k==6||k==9, return(k)); if(vecsearch(Vec(v), k), return(sumdigits(abs(k)))); listput(v, k)); if(k<0, k+=sumdigits(-k)^2; c+=1; if(k==2||k==3||k==0||k==6||k==9, return(k)); if(vecsearch(Vec(v), k), return(sumdigits(abs(k)))); listput(v, k)); c+=1)
for(n=1,10^4,if(a(n)==0,print1(n,", ")))

A290223 Algorithm: s(k) = n. s(k+1) = s(k) - digitsum(s(k))^2 if s(k) >= 0 and s(k+1) = s(k)+digitsum(abs(s(k)))^2 if s(k) < 0. Below gives the end behavior for each number n.

Original entry on oeis.org

0, 2, 3, 6, 6, 6, 3, 11, 9, 9, 3, 3, 6, 6, 6, 3, 11, 9, 0, 3, 3, 6, 2, 6, 3, 11, 9, 9, 11, 3, 6, 3, 6, 3, 11, 9, 9, 11, 3, 6, 3, 6, 3, 6, 9, 9, 11, 3, 6, 3, 6, 3, 6, 9, 9, 11, 3, 6, 6, 6, 3, 2, 9, 9, 11, 3, 6, 6, 6, 3, 3, 9, 9, 11, 3, 6, 2, 6, 3, 3, 0, 9, 11, 3, 6, 6, 6, 3, 6, 9, 9, 11, 3, 6, 6, 6, 3, 6, 9, 9, 3, 3, 6, 3, 6, 3, 3, 9, 9

Offset: 1

Derek Orr, Jul 24 2017

0 means the sequence s(k) becomes the 0 sequence.
2 means the sequence s(k) becomes 2, -2, 2, -2, ...
3 means the sequence s(k) becomes 3, -6, 30, 21, 12, 3, ...
6 means the sequence s(k) becomes 6, -30, -21, -12, -3, 6, ...
9 means the sequence s(k) oscillates between two numbers, each of which have a digit sum of 9. For example, 18 -> -63 -> 18 -> -63 -> ... so a(18) = 9.
11 means the sequence s(k) oscillates between two numbers, each of which have a digit sum of 11. For example, 65 -> -56 -> 65 -> ... so a(65) = 11.
a(n) = 2 for n = 2, 23, 62, 77, 119, 194, 287, 398. The next number n such that a(n) = 2 is over 10^5. This is believed to be finite.
a(n) = 11 for n = 8, 17, 26, 29, 35, 38, 47, 56, 65, 74, 83, 92, 149, 158, 167, 197. The next number n is over 10^5. This is believed to be finite.
The subsequences when a(n)=0, 3, 6, and 9 are believed to be infinite.

			a(19) = 0 because 19 - (1+9)^2 = -81. Then -81 + (8+1)^2 = 0.
a(13) = 6 because 13 - (1+3)^2 = -3. Then -3 + (3)^2 = 6.
a(17) = 11 because 17 - (1+7)^2 = -47. Then -47 + (4+7)^2 = 74. Then 74 - (7+4)^2 = -47, and so on.
a(23) = 2 because 23 - (2+3)^2 = -2. Then -2 + (2)^2 = 2.
a(25) = 3 because 25 - (2+5)^2 = -24. Then -24 + (2+4)^2 = 12. Then 12 - (1+2)^2 = 3.
a(28) = 9 because 28 - (2+8)^2 = -72. Then -72 + (7+2)^2 = 9. Then 9-(9)^2 = -72, and so on.

Cf. A007953, A003132.

a(n)=k=n;c=1;v=List();listput(v,k);while(c,if(k>=0,k-=sumdigits(k)^2;c+=1;if(k==2||k==3||k==0||k==6||k==9,return(k));if(vecsearch(Vec(v),k),return(sumdigits(abs(k))));listput(v,k));if(k<0,k+=sumdigits(-k)^2;c+=1;if(k==2||k==3||k==0||k==6||k==9,return(k));if(vecsearch(Vec(v),k),return(sumdigits(abs(k))));listput(v,k));c+=1)

A260465 a(n) is the smallest number not already in the sequence such that a(n)^3 begins with n.

Original entry on oeis.org

1, 3, 7, 16, 8, 4, 9, 2, 21, 10, 48, 5, 11, 52, 25, 55, 12, 57, 27, 59, 6, 61, 62, 29, 63, 64, 14, 66, 31, 67, 68, 32, 15, 70, 33, 154, 72, 73, 34, 74, 161, 35, 76, 164, 77, 36, 78, 169, 17, 37, 80, 174, 81, 38, 82, 178, 83, 18, 39, 182, 85, 184, 86, 40, 87, 188, 189, 19, 191, 89, 193, 90, 194, 42, 91, 197, 92, 199, 43

Offset: 1

Derek Orr, Jul 26 2015

Conjectured to be a permutation of the natural numbers.

Derek Orr, Table of n, a(n) for n = 1..10000

Cf. A018852.

v=[]; k=1; while(#v<100, d=digits(k^3); D=digits(#v+1); if(#D<=#d, c=1; for(i=1, #D, if(D[i]!=d[i], c=0; break)); if(c&&!vecsearch(vecsort(v), k), v=concat(v, k); k=0)); k++); v

a(n) >= n^(1/3) for all n > 0. If a(n) = n^(1/3), then n is a cube. Note the converse is false: a(27) = 14.

A260463 a(n) is the smallest number not already in the sequence such that a(n)^2 begins with n.

Original entry on oeis.org

1, 5, 6, 2, 23, 8, 27, 9, 3, 10, 34, 11, 37, 12, 39, 4, 42, 43, 14, 45, 46, 15, 48, 49, 16, 51, 52, 17, 54, 55, 56, 18, 58, 59, 188, 19, 61, 62, 63, 20, 203, 65, 66, 21, 213, 68, 69, 22, 7, 71, 72, 229, 73, 74, 235, 75, 24, 241, 77, 78, 247, 25, 251, 80, 81, 257, 26, 83, 263, 84, 267, 85, 86, 273, 87, 276, 88, 28, 89

Offset: 1

Derek Orr, Jul 26 2015

Conjectured to be a permutation of the natural numbers.

Differs from A018851 at n = 25.

Derek Orr, Table of n, a(n) for n = 1..10000

Cf. A018851.

v=[];k=1;while(#v<100,d=digits(k^2);D=digits(#v+1);if(#D<=#d,c=1;for(i=1,#D,if(D[i]!=d[i],c=0;break));if(c&&!vecsearch(vecsort(v),k),v=concat(v,k);k=0));k++);v

a(n) >= sqrt(n) for all n > 0. If a(n) = sqrt(n), then n is a square. Note the converse is false: a(25) = 16.

cp's OEIS Frontend

User: Derek Orr

A339928 Numbers k such that the removal of all terminating even digits from k! leaves a prime.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

PARI

Python

A338941 a(1)=1. For n >= 2, let S be the sum of all prime digits in a(1), a(2), ... a(n-1) and let C be the next nonprime number not already in the sequence. If S is a prime less than C and is not already a term of the sequence, a(n) = S. Otherwise, a(n) = C.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

PARI

A338938 a(1)=0. For n >= 2, let S be the sum of all nonprime digits in a(1), a(2), ... a(n-1) and let P be the next prime not already in the sequence. If S is a nonprime number less than P and not already in the sequence, a(n) = S. Otherwise, a(n) = P.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

PARI

A338921 a(0)=1, a(n) for n >= 1 is the number of distinct sums of two elements in [a(0), ..., a(n-1)], chosen without replacement.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Maple

Mathematica

PARI

A290227 Numbers n such that A290223(n) = 11.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

PARI

A290226 Numbers n such that A290223(n) = 2.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

PARI

A290224 Numbers n such that A290223(n) = 0.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

PARI

A290223 Algorithm: s(k) = n. s(k+1) = s(k) - digitsum(s(k))^2 if s(k) >= 0 and s(k+1) = s(k)+digitsum(abs(s(k)))^2 if s(k) < 0. Below gives the end behavior for each number n.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

PARI

A260465 a(n) is the smallest number not already in the sequence such that a(n)^3 begins with n.

Views

Author