Christian N. K. Anderson - cp's OEIS frontend

A375098 Diagonals of a Euclidian solid such that there exists a Pythagorean quadruple d^2=a^2+b^2+c^2 that is more cube-like than any prior value of d.

3, 9, 11, 41, 123, 153, 571, 1713, 2131, 7953, 23859, 29681, 110771, 332313, 413403, 1542841, 4628523, 5757961, 21489003, 64467009, 80198051, 299303201, 897909603, 1117014753, 4168755811, 12506267433, 15558008491, 58063278153, 174189834459, 216695104121

Offset: 1

Christian N. K. Anderson, Mark K. Transtrum, and David D. Allred, Jul 29 2024

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To determine how "cube-like" a Pythagorean quad is, we use the quotient C/(C-a*b*c) where C is the volume of an ideal cube for a given diagonal d, C=(d/sqrt(3))^3. A ratio of 100 indicates that the best {a,b,c} combination creates a solid 1 part per 100 smaller than the ideal cube volume.
Contains all the terms in A001835 and A079935 except the leading 1s. For such a term b(m) contained in a(n) from those sequences, 2*b(m) will tie the current record, while 3*b(m) will tie but will also break the current record exactly half the time and thus appear as a(n+1). This means round((2+sqrt(3))*b(m)) will also be in the sequence as either a(n+1) or a(n+2) if a better quad for 3*b(n) was found.
The constant (2+sqrt(3)) follows from the recurrence relationship b(n)=4*b(n-1)-b(n-2). The constant can be represented as the continued fraction 3,1,2,1,2,1,2,1,2,.... The equivalents for the 2D case (A001653) are 3+2*sqrt(2) and {5,1,4,1,4,1,4,1,4,...}.
We conjecture this method provides complete solutions. This was confirmed directly by brute-force testing up to 1e7 (optimized using the sum-of-two-squares theorem for a given d^2-a^2 = b^2 + c^2; see link below).

			3 is in the sequence because 3^2=1^2+2^2+2^2 is the smallest Pythagorean quad, with an error of one part in 4.344.
6 is NOT in the sequence because {6,2,4,4} is the most cube-like Pythagorean quad, but only ties the previous record without breaking it.
7 is NOT in the sequence because the most cube-like quad {7,2,3,6} has an error of one part in 2.2, worse than that for d=3.
9 is in the sequence NOT because of {9,3,6,6} which ties the previous record, but because {9,4,4,7} improves on the previous record with an error of one part in 4.958.

Christian N. K. Anderson, Table of n, a(n) for n = 1..1000
Christian N. K. Anderson, Table of most cube-like Pythagorean quadruples and the corresponding and the error quotient for d=1..10000
Christian N. K. Anderson, Table of d, a, b, c and error quotient for n = 1..50.
Christian N. K. Anderson, Mathematica code for generating this sequence with different amounts of rigorous error checking.

Cf. A096907, A096908, A096909, A096910 for lists of a, b, c, and d of the 10,000 first Pythgorean quads, sorted by ascending d.
Contains all terms in A001835 and A079935 except the leading 1s.
Cf. A001653: The 2D equivalent of this sequence (i.e., right triangle whose legs are closest to equal)

(* An efficient program is provided in the links section. *)

For n == 0 (mod 3), a(n) = 4*a(n-2)-a(n-3) OR a(n) = floor(a(n-1)*(2+sqrt(3))/3),
For n == 1 (mod 3), a(n) = 4*a(n-2)-a(n-1) OR a(n) = floor(a(n-1)*(2+sqrt(3))),
For n == 2 (mod 3), a(n) = 3*a(n-1).

A362681 The number of steps, starting from n, to reach x<=2 in an iteration x <- 2x - {sum of proper factors of 2x}.

Original entry on oeis.org

0, 0, 1, 1, 2, 1, 3, 1, 1, 1, 2, 1, 3, 1, 1, 1, 2, 1, 3, 1, 1, 3, 2, 1, 2, 4, 1, 1, 2, 1, 3, 1, 1, 3, 1, 1, 2, 4, 1, 1, 2, 1, 3, 1, 1, 3, 2, 1, 5, 1, 1, 1, 2, 1, 3, 1, 1, 3, 2, 1, 3, 3, 1, 1, 2, 1, 3, 2, 1, 1, 2, 1, 3, 4, 1, 3, 2, 1, 3, 1, 1, 2, 2, 1, 3, 3, 1, 1

Offset: 1

Kevin L. Schwartz and Christian N. K. Anderson, May 01 2023

A proper factor is defined as any divisor of n other than 1 and itself (Derbyshire).
The iteration step is x <- A157449(2x).
The iteration ends on the step after reaching half of any abundant number A005101/2.
a(1682)=7 is the only number over 6 in the first 10^6 terms.
Powers of 2 reach 2 in the first step, and then would enter an infinite loop if the iteration ended only when x <= 1.

J. Derbyshire, Prime Obsession: Bernhard Riemann and the Greatest Unsolved Problem in Mathematics. Penguin, 2004, p. 32.

Christian N. K. Anderson, Table of n, a(n) for n = 1..10000
Christian N. K. Anderson, Graph showing sparsity of 6s

Cf. A157449, A005101, A362684 (indices of records).

a(n) = my(ret=0); while(n>2, n = 4*n+1-sigma(2*n); ret++); ret; \\ Kevin Ryde, May 09 2023

A362684 a(n) is the index at which n first occurs in A362681.

Original entry on oeis.org

1, 3, 5, 7, 26, 49, 632, 1682

Offset: 0

Kevin L. Schwartz and Christian N. K. Anderson, May 01 2023

Among terms n=1..7, each number iterates through the previous terms. For example, the first iteration takes 1682 to 632=a(6), the second takes 632 to 49=a(5) and so forth.
1682 is the only number < 10^6 that requires 7 iterations to reach completion.
a(8), if it exists, is larger than 10^6.
a(8) > 10^9. - Michel Marcus, Oct 11 2023

			a(7) = 1682 because the first number to require 7 iterations to reach 2 or less is 1682.
Each iteration step is n <- 2n - {sum of proper factors of 2n} = A157449(2n); for n = 1682, this gives 2n = 3364, 3364 - (2 + 4 + 29 + 58 + 116 + 841 + 1682) = 632, which is a(6).
Note that the proper factors of x are all divisors of x other than 1 and x.

Cf. A362681, A157449.

Cf. A048133.

a(0)=1 prepended by Michel Marcus, Oct 11 2023

A226296 The smallest number beginning with n that can be decomposed into divisors consisting exclusively of the first n semiprimes.

Original entry on oeis.org

16, 24, 3456, 466560, 5927040, 6350400, 771573600, 838252800, 990186120000, 102979356480000, 112378266000000, 12379589782560000, 133730136540000000, 140865976625774400000, 1553794408841875200000, 16774637597496979200000, 17868635701681564800000

Offset: 1

Jonathan Vos Post, Christian N. K. Anderson, and Kevin L. Schwartz, Jun 02 2013

This is to A225903 as semiprimes A001358 are to prime A000040, and as Product of the first n semiprimes A112141 is to Primorials A002110.

			a(1) = 16, the smallest multiple of the first semiprime (4) that begins with leftmost digit 1.
a(2) = 24 = 4 * 6, the product of the first 2 semiprimes, and already begins with 2.

Christian N. K. Anderson, Table of n, a(n) for n = 1..375
Christian N. K. Anderson, Decomposition of the first 376 terms, indicating semiprime factors whose exponents are greater than one.

Cf. A001358, A112141, A225903.

A226531 Cubes that become prime when their least-significant (rightmost) digit is removed.

Original entry on oeis.org

27, 3375, 4096, 4913, 35937, 97336, 110592, 148877, 421875, 681472, 1191016, 1442897, 1560896, 2628072, 3241792, 3581577, 3869893, 4741632, 5359375, 8998912, 10218313, 12649337, 16777216, 16974593, 21253933, 26730899, 31255875, 32157432, 43986977, 45882712

Offset: 1

Kevin L. Schwartz and Christian N. K. Anderson, Jun 09 2013

			4096 = 16^3, and becomes the prime number 409 when truncated.

Christian N. K. Anderson, Table of n, a(n) for n = 1..10000

Cf. A226090.

Cf. A024770, A225873, A225885, A226354.

Select[Range[1000]^3,PrimeQ[Floor[#/10]]&] (* Harvey P. Dale, May 28 2021 *)

library(gmp)trimR=function(x) { x=as.character(x); ifelse(nchar(x)<2,0,substr(x,1,nchar(x)-1)) }
y=as.bigz(rep(0,10000)); len=0; n=as.bigz(-1)
while(len<10000) if(isprime(trimR((n=n+1)^3))) y[(len=len+1)]=n^3

A226087 Number of values k in base n for which the sum of digits of k = sqrt(k).

Original entry on oeis.org

1, 4, 2, 3, 3, 6, 2, 2, 2, 5, 2, 6, 2, 5, 5, 2, 2, 4, 2, 6, 6, 4, 2, 5, 2, 4, 2, 6, 2, 11, 2, 2, 6, 4, 5, 6, 2, 4, 6, 5, 2, 11, 2, 6, 5, 4, 2, 6, 2, 4, 6, 5, 2, 4, 5, 5, 6, 4, 2, 13, 2, 4, 4, 2, 5, 11, 2, 5, 6, 11, 2, 5, 2, 4, 6, 6, 6, 11, 2, 5, 2, 4, 2, 12, 5

Offset: 2

Christian N. K. Anderson and Kevin L. Schwartz, May 25 2013

Values of k in base n have at most 3 digits. Proof: Because sqrt(k) increases faster than the digit sum of k, only numbers with d digits meeting the condition d*(n-1) >= n^(d/2) are candidate fixed points. d < 3 for n > 6, and since there are no fixed points of four or more digits in bases 2 through 5, there are no fixed points in any base with more than 3 digits.

From the above, it can be shown that for three-digit fixed points of the form xyz, x <= 6; also x <= 4 for n > 846. These theoretical upper limits are statistically unlikely, and in fact of the 86356 solutions in bases 2 to 10000, only 6.5% of them begin with 2, and none begin with 3 through 6.

			For a(16)=5 the solutions are the square numbers {1, 36, 100, 225, 441} because in base 16 they are written as {1, 24, 64, E1, 1B9} and
  sqrt(1) = 1
  sqrt(36) = 6 = 2+4
  sqrt(100) = 10 = 6+4
  sqrt(225) = 15 = 14+1, and
  sqrt(441) = 21 = 1+11+9.

Christian N. K. Anderson, Table of n, a(n) for n = 2..10000

Cf. A226224.

Cf. digital sums for digits at various powers: A007953, A003132, A055012, A055013, A055014, A055015.

sapply(2:16,function(n) sum(sapply((1:(n^ifelse(n>6,1.5,2)))^2, function(x) sum(inbase(x,n))==sqrt(x))))

A226354 Squares that become cubes when their rightmost digit is removed.

Original entry on oeis.org

1, 4, 9, 16, 81, 10000, 640000, 7290000, 40960000, 156250000, 188210961, 466560000, 1176490000, 2621440000, 5314410000, 10000000000, 17715610000, 29859840000, 48268090000, 75295360000, 113906250000, 167772160000, 241375690000, 340122240000, 470458810000

Offset: 1

Christian N. K. Anderson and Kevin L. Schwartz, Jun 04 2013

			188210961=13719^2, while 18821096=266^3.

Christian N. K. Anderson, Table of n, a(n) for n = 1..500
Christian N. K. Anderson, Decomposition of a(n) into its square root, and truncated cube root for n=0..500.

Cf. A024770, A225873, A225885.

cQ[n_]:=IntegerQ[Surd[FromDigits[Most[IntegerDigits[n]]],3]]; Select[Range[ 700000]^2,cQ] (* Harvey P. Dale, Feb 21 2014 *)

trimR=function(x) { x=as.character(x); ifelse(nchar(x)<2,0,substr(x,1,nchar(x)-1)) }
iscube<-function(x) ifelse(as.bigz(x)<2,T,all(table(as.numeric(factorize(x)))%%3==0))
which(sapply(1:6400,function(x) iscube(trimR(x^2))))^2

For n > 11: a(n)=(100*(n-6)^3)^2 (188210961 is the last "exception" as is easy to prove with the help of the Nagell-Lutz theorem). - Reiner Moewald, Dec 30 2013

A226352 Number of integers k in base n whose squared digits sum to sqrt(k).

Original entry on oeis.org

1, 3, 2, 2, 1, 1, 4, 2, 1, 2, 3, 6, 1, 6, 3, 3, 1, 2, 2, 3, 2, 4, 4, 4, 2, 9, 2, 4, 2, 3, 1, 3, 3, 3, 3, 1, 2, 4, 5, 4, 1, 6, 1, 5, 2, 5, 2, 5, 4, 1, 3, 5, 1, 5, 2, 5, 1, 7, 3, 2, 2, 7, 3, 2, 2, 4, 3, 2, 1, 3, 3, 6, 3, 3, 2, 1, 2, 5, 3, 4, 1, 4, 1, 3, 2, 3, 1

Offset: 2

Christian N. K. Anderson and Kevin L. Schwartz, Jun 04 2013

Any d-digit number in base n meeting the criterion must also meet the condition d*(n-1)^2 < n^(d/2). Numerically, it can be shown this limits the candidate values to squares < 22*n^4. The larger values are statistically unlikely, and in fact the largest value of k in the first 1000 bases is ~9.96*n^4 in base 775.

			In base 8, the four solutions are the values {1,16,256,2601}, which are written as {1,20,400,5051} in base 8 and
sqrt(1)    =  1 = 1^2;
sqrt(16)   =  4 = 2^2 + 0^2;
sqrt(256)  = 16 = 4^2 + 0^2 + 0^2;
sqrt(2601) = 51 = 5^2 + 0^2 + 5^2 + 1^2,

Christian N. K. Anderson, Table of n, a(n) for n = 2..1000

Cf. A226353.
Cf. A226353, A226224.
Cf. digital sums for digits at various powers: A007953, A003132, A055012, A055013, A055014, A055015.

inbase=function(n, b) { x=c(); while(n>=b) { x=c(n%%b, x); n=floor(n/b) }; c(n, x) }
for(n in 2:50) cat("Base", n, ":", which(sapply((1:(4.7*n^2))^2, function(x) sum(inbase(x, n)^2)==sqrt(x)))^2, "\n")

A226353 Largest integer k in base n whose squared digits sum to sqrt(k).

Original entry on oeis.org

1, 49, 169, 36, 1, 1, 2601, 1089, 1, 8836, 33489, 44100, 1, 149769, 128164, 96721, 1, 156816, 1225, 40804, 12321, 831744, 839056, 1149184, 1737124, 3655744, 407044, 1890625, 2208196, 1089, 1, 1466521, 6125625, 2235025, 2832489, 1, 3759721, 6885376, 8844676

Offset: 2

Christian N. K. Anderson and Kevin L. Schwartz, Jun 04 2013

Any d-digit number in base n meeting the criterion must also meet the condition d*(n-1)^2 < n^(d/2). Numerically, it can be shown this limits the candidate values to squares < 22*n^4. The larger values are statistically unlikely, and in fact the largest value of k in the first 1000 bases is ~9.96*n^4 in base 775.

a(n)=1 iff A226352(n)=1.

			In base 8, the four solutions are the values {1,16,256,2601}, which are written as {1,20,400,5051} in base 8 and
sqrt(1)   = 1  = 1^2
sqrt(16)  = 4  = 2^2+0^2
sqrt(256) = 16 = 4^2+0^2+0^2
sqrt(2601)= 51 = 5^2+0^2+5^2+1^2

Christian N. K. Anderson, Table of n, a(n) for n = 2..1000
Christian N. K. Anderson, Table of base, all solutions in base 10, and all solutions in base n for bases 2 to 1000.

Cf. A226352, A226224.

Cf. digital sums for digits at various powers: A007953, A003132, A055012, A055013, A055014, A055015.

inbase=function(n,b) { x=c(); while(n>=b) { x=c(n%%b,x); n=floor(n/b) }; c(n,x) }
for(n in 2:50) cat("Base",n,":",which(sapply((1:(4.7*n^2))^2,function(x) sum(inbase(x,n)^2)==sqrt(x)))^2,"\n")

A226224 The largest value of k in base n for which the sum of digits of k = sqrt(k).

Original entry on oeis.org

1, 25, 9, 64, 100, 144, 49, 64, 81, 225, 121, 441, 169, 441, 441, 256, 289, 324, 361, 1296, 1296, 484, 529, 1089, 625, 676, 729, 2401, 841, 2601, 961, 1024, 3025, 1156, 2500, 4096, 1369, 1444, 4356, 3136, 1681, 4900, 1849, 5929, 3025, 2116, 2209, 6561, 2401

Offset: 2

Christian N. K. Anderson and Kevin L. Schwartz, May 31 2013

There are no values of k in base n with more than 3 digits. Proof: such a value with d digits would need to meet the criterion d*(n-1)>=sqrt(n)^d which establishes an upper limit of 4 digits for 2<=n<=6 and 3 for n>6. Because there are no four digit values of k in bases 2 through 6, k has a maximum of three digits in all bases.
Because k must be a square, there are only sqrt(n)^3 possible values in any base.
From the above, it can be shown that for three-digit fixed points of the form xyz, x <= 6; also x<=4 for n>846. These theoretical upper limits are statistically unlikely, and in fact of the 86356 solutions in bases 2 to 10000, only 6.5% of them begin with 2, and none begin with 3 through 6.
a(n)=1 iff A226087(n)=1. Conjecture: this occurs exactly once -- in base 2.

			For a(16) the solutions are the square numbers {1, 36=6^2, 100=10^2, 225=15^2, 441=21^2} because in base 16 they are written as {1, 24, 64, E1, 1B9} and 1 = 1, 6 = 2+4, 10 = 6+4, 15 = 14+1, and 21 = 1+11+9.

Christian N. K. Anderson, Table of n, a(n) for n = 2..10000
Christian N. K. Anderson, Table of base, max(k), number of k, and all values of k for base 2 to 10000.

Cf. A007953, A003132, A055012, A055013, A055014, A055015, A226087.

for(n in 2:500) cat("Base",n,":",which(sapply((1:(ifelse(n>6,7,1)*n^ifelse(n>6,1,2)))^2, function(x) sum(inbase(x,n))==sqrt(x)))^2, "\n")

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User: Christian N. K. Anderson

A375098 Diagonals of a Euclidian solid such that there exists a Pythagorean quadruple d^2=a^2+b^2+c^2 that is more cube-like than any prior value of d.

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A362681 The number of steps, starting from n, to reach x<=2 in an iteration x <- 2x - {sum of proper factors of 2x}.

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A362684 a(n) is the index at which n first occurs in A362681.

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A226296 The smallest number beginning with n that can be decomposed into divisors consisting exclusively of the first n semiprimes.

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A226531 Cubes that become prime when their least-significant (rightmost) digit is removed.

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A226087 Number of values k in base n for which the sum of digits of k = sqrt(k).

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A226354 Squares that become cubes when their rightmost digit is removed.

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A226352 Number of integers k in base n whose squared digits sum to sqrt(k).

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